Let $(X_n)_n$ a sequence of independent random variables such that $X_1=0$ and for all $j≥2$:
$$ k \mapsto \mathbb{P}\left(X_{j}=k\right)= \begin{cases}\frac{1}{j^{3}} & k \in\{ \pm 1, \pm 2, \ldots, \pm j\} \\ 1-\frac{2}{j^{2}} & k=0\end{cases} $$
Show that $\displaystyle\frac{1}{n^{\alpha}} \sum_{j=1}^{n} X_{j} \stackrel{\mathbb{P}}{\longrightarrow} 0$ para $\left.\alpha \in\right] \frac{1}{2}, \infty[$.
My attempt: Let $\varepsilon>0$, $\alpha\in(\frac{1}{2},\infty)$ and $n\in \mathbb{N}$. Note that by Markov inequality we have that:
\begin{align} \mathbb{P}\left(\frac{1}{n^{\alpha}}\left|\sum_{j=1}^{n}X_{j}\right|>\varepsilon\right)=\mathbb{P}\left(\frac{1}{n^{2\alpha}}\left(\sum_{j=1}^{n}X_{j}\right)^{2}>\varepsilon^2\right)\leq \frac{1}{\varepsilon^2}\mathbb{E}\left(\frac{1}{n^{2\alpha}}\left(\displaystyle\sum_{j=1}^{n}X_{j}\right)^{2}\right) \end{align} Note that $\mathbb{E}(X_{1})=0$. Now for $j>1$ we have that: $$\mathbb{E}(X_{j})=\sum_{k=-j}^{j}k\cdot\mathbb{P}(X_{j}=k)=-j\frac{1}{j^3}-(j-1)\frac{1}{j^3}-\ldots+(j-1)\frac{1}{j^3}+j\frac{1}{j^3}=0$$ Let us considerer $M_{X_{j}}(t)$, with $t\in \mathbb{R}$, the moment generating function of $X_{j}$, which is: $$M_{X_{j}}(t)=\mathbb{E}(e^{tX_{j}})=\sum_{k=-j}^{j}e^{tk}\cdot\mathbb{P}(X_{j}=k)=\sum_{k=-j}^{-1}e^{tk}\frac{1}{j^3}+\left(1-\frac{2}{j^2}\right)+\sum_{k=1}^{j}e^{tk}\frac{1}{j^3}$$ In this way, we obtain that: $$\mathbb{E}(X_{j}^2)=\frac{1}{j^3}\sum_{k=-j}^{j}k^2=\frac{(j+1)(2j+1)}{3j^2}$$ Observe that since the variables $(X_n)$ are independent, it follows inductively that: $$\mathbb{E}\left(\left(\sum_{j=1}^{n}X_{j}\right)^2\right)=\sum_{j=1}^{n}\mathbb{E}(X_{j}^2)$$ Given that $\alpha\in \left]\frac{1}{2},\infty\right[$ we have the following: \begin{align} \frac{1}{\varepsilon^2}\mathbb{E}\left(\frac{1}{n^{2\alpha}}\left(\displaystyle\sum_{j=1}^{n}X_{j}\right)^{2}\right)=\frac{1}{\varepsilon^2 n^{2\alpha}}\sum_{j=1}^{n}\mathbb{E}(X_{j}^2)=\frac{1}{\varepsilon^2 n^{2\alpha}}\sum_{j=1}^{n}\frac{(j+1)(2j+1)}{3j^2}\underset{n\to \infty}{\longrightarrow} 0 \end{align} So we conclude that $\displaystyle\frac{1}{n^{\alpha}} \sum_{j=1}^{n} X_{j} \stackrel{\mathbb{P}}{\longrightarrow} 0$
Is this development correct?